Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `pi/4`, and the triangle's area is `12`. What is the area of the triangle's incircle?

Answer 1

`color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.0444^2 = 0.0062`

Explanation:

The distances from the incenter to each side are equal to the inscribed circle's radius. The area of the triangle is equal to 12×r×(the triangle's perimeter), 1 2 × r × ( the triangle's perimeter ) , where r is the inscribed circle's radius.

First to find the perimeter of the triangle.

`hat A = pi/12, hat B = pi/4, hat C = (2pi)/3, A_t = 12`

Area of triangle `A_t = (1/2)ab sin C = (1/2)bc sin A = (1/2)ca sin B`

`ab = (12 * 2) / sin ((2pi)/3) = 27.71`

Similarly, `bc = (12 * 2) / sin (pi/12) = 92.73`

`ca = (12 * 2) / sin (pi/4) = 33.91`

`ab * bc * ca = (abc)^2 = 27.71 * 92.73 * 33.91`

`abc = sqrt(27.71 * 92.73 * 33.91) = 295.18`

`a = (abc) / (bc) = 295.18 / 92.73 = 3.18`

Likewise, `b = (abc) / (ca) = 295.18 / 33.91 = 8.7`

`c = (abc) / (ab) = 295.18 / 27.71 = 10.65`

`"Perimeter of the triangle " p = a + b + c = 3.18 + 8.7 + 10.65 = 22.53`

`"Radius of incircle " r = A_t / (12 * p) = 12 / (12 * 22.53) = 0.0444`

`color(purple)("Area of inscribed circle " A_r = pi r^2 = pi * 0.0444^2 = 0.0062`